1) The axial vector current $j^{\mu 5}$ is a pseudovector
$$j^{\mu 5}~:=~\overline{\psi}\gamma^{\mu}\gamma^5\psi~=~j^{\mu}_R-j^{\mu}_L,\qquad
j^{\mu}_{R,L}~:=~ \overline{\psi}_{R,L}\gamma^{\mu}\psi_{R,L}, $$
$$\psi_{R,L}~:=~P_{R,L}\psi,\qquad P_{R,L} ~:=~\frac{1\pm\gamma^5}{2} . $$
The $4$-divergence $d_{\mu}j^{\mu 5}$ is a pseudoscalar. That the axial current $j^{\mu 5}$ is conserved classically means that the $4$-divergence $d_{\mu}j^{\mu 5}=0$ vanishes classically, and if one defines the axial charge
$$N^5(t)~:=~N_R(t)-N_L(t),\qquad N_{R,L}(t)~:=~\int {\rm d}^3{x}~j^0_{R,L}(t,\vec{x}),$$
then $N^5(t)$ is conserved over time classically.
2) It follows from the Dirac equation that a spin $1/2$ particle and its antiparticle must have opposite intrinsic parity. Conventionally, for quarks $P(q)=1=-P(\bar{q})$.
Thus the parity of a meson is
$$P({\rm meson})~=~P(q)P(\bar{q})(-1)^{L}~=~(-1)^{L+1}.$$
In particular, a pion $\pi^0$ with $J=L=S=0$ is a pseudoscalar, with parity $P(\pi^0)=-1$.
3) A pion is a bound state of a quark and an antiquark, which is difficult to directly relate to the Lagrangian density of the standard model, and ultimately to the two photons $\gamma+\gamma$. In practice, one instead studies how the $\pi^0$ and the two $\gamma's$ couple to the axial vector current $j^{\mu 5}$.
Quoting Peskin and Schroeder on the bottom of page 669: We can parametrize the matrix element of $j^{\mu5a}$ between the vacuum and an on-shell pion by writing
$$ \langle 0 | j^{\mu5a}(x) | \pi^b(p) \rangle~=~ -i p^{\mu} f_{\pi} \delta^{ab} e^{-ip\cdot x}, \qquad (19.88) $$
where $a,b$ are isospin indices and $f_{\pi}$ is a constant [...]. As a consistency check of eq. (19.88), note that lhs = pseudovector $\times$ pseudoscalar=vector=rhs.
On the other hand, it is e.g. argued in Chapter 76 of Srednecki, QFT, via a LSZ formula and a Ward identity, that the $4$-divergence
$$d_{\mu} \langle p,q | j^{\mu5}(x) | 0 \rangle \qquad \qquad \qquad (76.20) $$
vanishes classically, where $\langle p,q |$ is a state with two outgoing photons with $4$-momenta $p$ and $q$.
So in a nutshell, the pion decay $\pi^0\to \gamma+\gamma$ is classically forbidden because a photon two-state doesn't couple classically to the axial current $j^{\mu5}$.
Looks like the classic "catch my sloppiness" exercise on Schwartz. (My students got extra credit for those). I edited your question to drop P&S in favor of S, clearly your intention. Let's only deal with $F_\pi\sim 93$MeV, to avoid confusion. In your (1), you took his τs to be Pauli σs, when he clearly takes them to be the real SU(2) generators, so σ/2 s, for his (28.26)... so the last member of your (1) is flawed.
Otherwise your (5) and (7) are, indeed, correct as conjectured. You only need compare normalizations for the neutral versus charged pion, so you could be cavalier about the absolute normalizations of currents!
Indeed, Matt almost certainly means
$J^5_{\mu , +}= J^5_{\mu 1,1} +i J^5_{\mu 1,2} \propto F_\pi \partial_\mu (\pi_1 + i \pi_2) + O(\pi^2)$ in his footnote.
You could convince yourself the axial current is $\propto \Sigma^\dagger \partial_\mu \Sigma - \Sigma \partial_\mu \Sigma ^\dagger\propto F_\pi \partial_\mu \pi^a \tau^a+ O(\pi^2)$ rewritten in 0, $\pm$ notation, as above, but I'll leave the "joy" for you.
The PDG thus uses $f_\pi= \sqrt {2} ~~ F_\pi $ ~ 130 MeV, as per Matt's footnote.
Best Answer
The dyslexia purgatory in the rabbit hole you've slipped into is "fields-versus-states". Weinberg, Scherer, and lots of purists don't discuss matrix elements, and hence fields and states; but, rather, neutral interaction terms in lagrangians. Your comment leaves what a "positively charged pion" is ambiguous: a field or a state?
In any case, the majority of users, like Weinberg, cf the definition of charged Ws here, use the convention you start with. As a result, the corresponding neutral interaction term of the σ model, i terms of fields, is $$ \propto \bar p \pi^+ n \propto \bar \psi \pi^+ \begin{pmatrix} 0&1\\0 &0\end{pmatrix} \psi , $$ visibly neutral, albeit counterintuitive, as you complained.
Few people will discuss the fact that it assumes the non- vanishing matrix element entering PCAC is $\langle 0| \pi^+|\pi^+\rangle$, i.e. $\pi^+|0\rangle \propto |\pi^-\rangle$, which would drive you (and me, and half of Schwartz, in case you noticed his inconsistency between (22.17) and (30.91)!) crazy...
They would rather take your (presumed favorite) $$ \pi^+|0\rangle \propto |\pi^+\rangle, $$ which, however, hides a hermitean conjugation in the interest of sanity. In this matrix element convention, all the +s and the -s add up to 0, so it is the matrix element
$$ \langle p| \pi^+|n\rangle \propto \langle p|p\rangle $$ which is neutral.
Much of the literature follows the leading, legit, convention, but lots of people stick to the second, and some just shrug the torpedoes off and assume the well-meaning reader will chase the signs... and wink at consistency...